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1/31/17
Gauss’s Law
A. B. Kaye, Ph.D.
Associate Professor of Physics
2 February 2017
Tentative Schedule
• Yesterday
•
Concluded our discussion of electric fields (Ch. 23)
• Today
•
•
Begin Gauss’s Law
First homework is DUE
• Tomorrow
•
Continue Gauss’s Law
GAUSS’S LAW
Introduction
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Gauss’s Law
• Gauss’s Law can be used as an alternative
procedure for calculating electric fields
• This law is based on the inverse-square
behavior of the electric force between point
charges
•
•
It is convenient for calculating the electric field of highly
symmetric charge distributions
Gauss’s Law is important in understanding and verifying
the properties of conductors in electrostatic equilibrium
GAUSS’S LAW
Electric Flux
Electric Flux
• The electric flux is the
product of the magnitude
of the electric field and the
surface area perpendicular
to the field:
• It has units of: N · m2 / C
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Electric Flux, General Area
• The electric flux is
proportional to the number
of electric field lines
penetrating some surface
• The field lines may make
some angle q with the
perpendicular to the
surface
• Therefore,
Electric Flux, Interpreting the Equation
• The flux is a maximum when the surface is
perpendicular to the field (i.e., when q = 0º)
• The flux is zero when the surface is parallel to
the field (i.e., when q = 90º)
• If the field varies over the surface, our relation
is valid for only a small element of the area
• Let’s try to create a more general expression…
Electric Flux, General
• In our more general case, let’s look at a
large surface that is divided into a large
number of small elements, DAi
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Electric Flux, Closed Surface
• Assume a closed
surface
• The vectors Ai point in
different directions
•
•
At each point, they are
perpendicular to the surface
By convention, they point
outward
• Let’s look at each one
of these in turn…
Flux Through Closed Surface, cont.
•
At (1), the field lines are
crossing the surface from the
inside to the outside;
q < 90o and F is positive
•
At (2), the field lines graze
surface; q = 90o and F = 0
•
At (3), the field lines are
crossing the surface from the
outside to the inside; 180º > q >
90º, and F is negative
Flux Through Closed Surface, final
• The net flux through the surface is
proportional to the net number of lines leaving
the surface
•
This net number of lines is the number of lines leaving the
surface minus the number entering the surface
• If En is the component of the field
perpendicular to the surface, then
•
The integral is over a closed surface
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Flux Through a Cube, Example
Consider a uniform electric
field E oriented in the x
direction in empty space
If a cube with edge length is
placed in the field (oriented as
shown), what is the net electric
flux through the surface of the
cube?
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